Question

use the long division method to find the result when $9x^3 - 3x^2 + 9x + 3$ is divided by $3x + 2$. if there is a remainder, express the result in the form $q(x) + \frac{r(x)}{b(x)}$.

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(9x^{3}\) by the leading term of the divisor \(3x\). So, \(\frac{9x^{3}}{3x}=3x^{2}\). This is the first term of the quotient \(q(x)\).
Multiply the divisor \(3x + 2\) by \(3x^{2}\): \(3x^{2}(3x + 2)=9x^{3}+6x^{2}\).
Subtract this from the dividend: \((9x^{3}-3x^{2}+9x + 3)-(9x^{3}+6x^{2})=9x^{3}-3x^{2}+9x + 3 - 9x^{3}-6x^{2}=-9x^{2}+9x + 3\).

Step2: Divide the new leading terms

Divide the leading term of \(-9x^{2}+9x + 3\) (which is \(-9x^{2}\)) by the leading term of the divisor \(3x\). So, \(\frac{-9x^{2}}{3x}=-3x\). This is the next term of the quotient.
Multiply the divisor \(3x + 2\) by \(-3x\): \(-3x(3x + 2)=-9x^{2}-6x\).
Subtract this from \(-9x^{2}+9x + 3\): \((-9x^{2}+9x + 3)-(-9x^{2}-6x)=-9x^{2}+9x + 3 + 9x^{2}+6x = 15x+3\).

Step3: Divide the new leading terms

Divide the leading term of \(15x + 3\) (which is \(15x\)) by the leading term of the divisor \(3x\). So, \(\frac{15x}{3x}=5\). This is the next term of the quotient.
Multiply the divisor \(3x + 2\) by \(5\): \(5(3x + 2)=15x+10\).
Subtract this from \(15x + 3\): \((15x + 3)-(15x+10)=15x + 3 - 15x - 10=-7\).

Answer:

The quotient \(q(x)=3x^{2}-3x + 5\) and the remainder \(r(x)=-7\) and the divisor \(b(x)=3x + 2\). So the result is \(3x^{2}-3x + 5+\frac{-7}{3x + 2}\) or \(3x^{2}-3x + 5-\frac{7}{3x + 2}\).